Consider a network path P-Q-R between nodes P and R via router Q. Node P sends a file of size 10^6 bytes to R via this path by splitting the file into chunks of 10^3 bytes each. Node P sends these chunks one after the other without any wait time between the successive chunk transmissions. Assume that the size of extra headers added to these chunks is negligible, and that the chunk size is less than the MTU. Each of the links P-Q and Q-R has a bandwidth of 10^6 bits/sec, and negligible propagation latency. Router Q immediately transmits every packet it receives from P to R, with negligible processing and queueing delays. Router Q can simultaneously receive on link P-Q and transmit on link Q-R. Assume P starts transmitting the chunks at time t = 0. Which one of the following options gives the time (in seconds, rounded to 3 decimal places) at which R receives all the chunks of the file?

To solve this, calculate total file size, chunk size, total chunks, chunk transmission time, and total end-to-end time considering pipelining. File size: 10^6 bytes = 8 * 10^6 bits. Chunk size: 10^3 bytes = 8 * 10^3 bits. Bandwidth: 10^6 bits/sec. Time to transmit one chunk: (8 * 10^3 bits) / (10^6 bits/sec) = 8 * 10^-3 seconds = 0.008 seconds. Total number of chunks: (10^6 bytes) / (10^3 bytes/chunk) = 1000 chunks. Since Q can simultaneously receive and transmit (pipelining), the second link's transmission starts almost immediately after the first chunk is received. The total time will be the time for the first chunk to reach R, plus the time for the remaining (N-1) chunks to be transmitted from P. Time for first chunk to reach R = Time to transmit from P-Q + Time to transmit from Q-R = 0.008s + 0.008s = 0.016s (if non-pipelined, but Q transmits immediately). In a pipelined scenario with 2 links, the bottleneck is the total time for all chunks to be sent from the source, plus the delay of the first chunk to traverse the network. Total time = Time for first chunk to reach R + (Total number of chunks - 1) * Time to transmit one chunk (from source). Time for first chunk to reach R: One chunk takes 0.008s to transmit on P-Q. Router Q receives it and immediately transmits it. So, another 0.008s to transmit on Q-R. The first chunk arrives at R at 0.008s + 0.008s = 0.016s. However, due to pipelining, the overall transmission is dominated by the total time P takes to send all chunks plus the last chunk's propagation. Total time P takes to transmit all chunks: 1000 chunks * 0.008s/chunk = 8.000 seconds. While P is sending chunk N, Q is sending chunk N-1. The last chunk (chunk #1000) starts transmission from P at time `(1000-1) * 0.008s = 7.992s`. It finishes transmission from P at `7.992s + 0.008s = 8.000s`. This chunk is then transmitted by Q. It finishes transmission from Q to R at `8.000s + 0.008s = 8.008s`. Therefore, R receives all chunks at 8.008 seconds. The correct option is B) 8.008

GATE 2024 CS Network: P-Q-R File Transfer Time Calculation Problem

Computer Sciences > GATE 2024 SET-1 > Networking

Consider a network path P-Q-R between nodes P and R via router Q. Node P sends a file of size 10⁶ bytes to R via this path by splitting the file into chunks of 10³ bytes each. Node P sends these chunks one after the other without any wait time between the successive chunk transmissions.
Assume that the size of extra headers added to these chunks is negligible, and that the chunk size is less than the MTU. Each of the links P-Q and Q-R has a bandwidth of 10⁶ bits/sec, and negligible propagation latency. Router Q immediately transmits every packet it receives from P to R, with negligible processing and queueing delays. Router Q can simultaneously receive on link P-Q and transmit on link Q-R. Assume P starts transmitting the chunks at time t = 0.
Which one of the following options gives the time (in seconds, rounded 3 to 3 decimal places) at which R receives all the chunks of the file?

8.000

8.008

8.016

16.000

Correct : c

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